55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
Answer:
1x10^-8 M
Explanation:
Since the solution turns blue, it mean the solution is a base.
Now, to know which option is correct, we need to determine the pH of each solution. This is illustrated below:
1. Concentration of Hydrogen ion, [H+] = 1x10^-2 M
pH =..?
pH = - log [H+]
pH = - log 1x10^-2
pH = 2
2. Concentration of Hydrogen ion, [H+] = 5x10-2 M
pH =..?
pH = - log [H+]
pH = - log 5x10^-2
pH = 1.3
3. Concentration of Hydrogen ion, [H+] = 5x10 M
pH =..?
pH = - log [H+]
pH = - log 5x10
pH = - 1.7
4. Concentration of Hydrogen ion, [H+] = 1x10-8 M
pH =..?
pH = - log [H+]
pH = - log 1x10^-8
pH = 8
A pH reading shows if the solution is acidic or basic. A pH reading between 0 and 6 indicates an acidic solution, a pH reading of 7 indicates a neutral solution while a pH reading between 8 and 14 indicates a basic solution.
From the above calculations, the pH reading indicates a basic solution when the hydrogen ion concentration was 1x10^-8 M.
The answer <span>is <span>8.9 g/mL</span>.</span>
The density (D) is <span>equal to mass (m) divided by volume (V): D = m/V
Let's find the mass of the object:
m = 156 g - 105.5 g = 50.5 g
Let's find the volume of the volume:
V = 30.7 mL - 25 mL = 5.7 mL
The density is:
D = m/V = 50.5 g / 5.7 mL = 8.9 g/mL</span>
Answer:
last choice
Explanation:
oxidation and reduction can be defined in terms of adding or removing oxygen to a compound
oxidation is gaining oxygen
reduction is to loss oxygen
Answer:
The sharing of electrons between a water molecule that forms four hydrogen bonds with the other four water molecules:
Explanation:
The hydrogen bond is a weak electrostatic force of attraction that exists between a covalently bonded H-atom and a highly electronegative atom like N,O or F.
In the case of the water molecule,
the highly electronegative atom is Oxygen and the intermolecular hydrogen bond in water is as shown below:
Thus H-bond is a weak electrostatic attraction formed between H-atom and O-atom in water.
